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Chapter 3: Problem 9

Show that if \(x_{n} \geq 0\) for all \(n \in \mathbb{N}\) and \(\lim \left(x_{n}\right)=0\), then \(\lim \left(\sqrt{x_{n}}\right)=0\).

Short Answer

Expert verified
By using the Sandwich theorem on the inequalities we derived from the given conditions, we've proven that the limit of sequence \(\sqrt{x_{n}}\) is indeed 0.

Step by step solution

01

Understand the Problem

We have \(x_{n} \geq 0\) for all \(n \in \mathbb{N}\) and given that \(\lim(x_{n}) = 0\). Our goal is to show that \(\lim(\sqrt{x_{n}}) = 0\).
02

Apply the Sandwich Theorem

Since we have \(x_{n} \geq 0\), we know that \(0 \leq \sqrt{x_{n}} \leq x_{n}\). This inequality holds because a nonnegative real number is greater than or equal to its square root. Furthermore, for \(x_{n}\), the limit as \(n\) approaches infinity goes to 0. Hence, \(\lim(x_{n}) = 0\). Since \(\sqrt{x_{n}}\) is sandwiched between \(x_{n}\) and \(0\), we can apply the Sandwich Theorem which says that if a function is squeezed between two other functions which both approach the same limit, then the squeezed function has also this limit.
03

Conclude the Proof

Therefore, from the Sandwich Theorem, we can say that \(\lim(\sqrt{x_{n}}) = 0\). This proves that if a sequence \(x_{n}\) of nonnegative real numbers converges to 0, then the sequence of their square roots also converges to 0.

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Most popular questions from this chapter

Show that (a) \(\lim \left(\frac{1}{\sqrt{n+7}}\right)=0\), (b) \(\lim \left(\frac{2 n}{n+2}\right)=2\), (c) \(\lim \left(\frac{\sqrt{n}}{n+1}\right)=0\), (d) \(\lim \left(\frac{(-1)^{n} n}{n^{2}+1}\right)=0\).

The sequence \(\left(x_{n}\right)\) is defined by che following formulas for the \(n\) th term. Write the furst five terms in each case: (a) \(x_{n}:=1+(-1)^{n}\), (b) \(x_{n}:=(-1)^{n} / n\), (c) \(x_{n}:=\frac{1}{n(n+1)}\), (d) \(x:=\frac{1}{n^{2}+2}\).

Show that if \(X\) and \(Y\) are sequences such that \(X\) and \(X+Y\) are convergent, then \(Y\) is convergent.

Determine the following limits. (a) \(\lim \left((3 \sqrt{n})^{1 / 2 n}\right)\), (b) \(\lim \left((n+1)^{1 / \ln (n+1)}\right)\).

Establish the convergence and find the limits of the following sequences. (a) \(\left((1+1 / n)^{n+1}\right)\) (b) \(\left((1+1 / n)^{2 n}\right)\) (c) \(\left(\left(1+\frac{1}{n+1}\right)^{n}\right)\), (d) \(\left((1-1 / n)^{n}\right)\)
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